Find the area of the parallelogram formed by the points $A ( 2 , - 3,5 ) , B ( 8 , - 2,7 )$, $C ( 3 , - 2,8 )$, and $D ( 9 , - 1,10 )$.
A) 282
B) $\sqrt { 10 }$
C) 10
D) $\sqrt { 282 }$
E) 13

Question

Quizplus · Accepted Answer

The area of a parallelogram formed by vectors $\vec{AB}$ and $\vec{AC}$ can be found using the cross product of these vectors, $\vec{AB} \times \vec{AC}$, and then finding the magnitude of this cross product. First, find vectors $\vec{AB}$ and $\vec{AC}$ by subtracting coordinates: $\vec{AB} = (8 - 2, -2 + 3, 7 - 5) = (6, 1, 2)$ and $\vec{AC} = (3 - 2, -2 + 3, 8 - 5) = (1, 1, 3)$. The cross product $\vec{AB} \times \vec{AC} = (1*3 - 2*1, 2*1 - 6*3, 6*1 - 1*1) = (3 - 2, 2 - 18, 6 - 1) = (1, -16, 5)$. The magnitude of this vector is $\sqrt{1^2 + (-16)^2 + 5^2} = \sqrt{1 + 256 + 25} = \sqrt{282}$, which gives the area of the parallelogram.

Find the Area of the Parallelogram Formed by the Points A(2,−3,5),B(8,−2,7)A ( 2 , - 3,5 ) , B ( 8 , - 2,7 )A(2,−3,5),B(8,−2,7)

Find the Area of the Parallelogram Formed by the Points $A ( 2 , - 3,5 ) , B ( 8 , - 2,7 )$